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DOUBLY INTERSPERSED SEQUENCES, DOUBLE INTERSPERSIONS,
AND FRACTAL SEQUENCES
CLARK KIMBERLING
Abstract. It is proved that distinct positive Fibonacci sequences eventually intersperse
or else doubly intersperse. Necessary and sufficient conditions for the latter are given. A
rectangular array of infinitely many rows is called a double interspersion if every pair of
its row sequences doubly intersperse. The union of the odd-numbered columns of a double
interspersion yield, as an order array, an interspersion, as does the union of even-numbered
columns. A special example of a double interspersion whose rows are Fibonacci sequences is
examined; the columns of the associated pair of interspersions all satisfy a certain 3rd-order
linear recurrence.
1. Introduction
We begin with four definitions. Sequences s = (s1 , s2 , . . .) and x = (x1 , x2 , . . .) eventually
intersperse if there exist i and j such that
0 < si < xj < si+1 < xj+1 < si+2 < xj+2 < · · · ;
(1.1)
that is, if si+h < xj+h for all h ≥ 0. Sequences s and x eventually doubly intersperse if there
exist i and j such that
0 < si < si+1 < xj < xj+1 < si+2 < si+3 < xj+2 < xj+3 < · · · .
(1.2)
If i is least possible in (1.2), then x doubly intersperses s from si and xj , and we write
D(s, x) = (i, j).
A Fibonacci sequence is a sequence s = (s1 , s2 , . . .) of positive real numbers (not necessarily
integers) satisfying
sm = sm−1 + sm−2
(1.3)
for m ≥ 3. The classical Fibonacci sequence (Fm ) is given by (1.3) using initial terms
F1 = s1 = 1 and F2 = s2 = 1.
The main result we shall prove is that each pair of distinct Fibonacci sequences eventually
intersperse or eventually doubly intersperse. We shall also examine arrays consisting of
infinitely many Fibonacci row sequences, every pair of which eventually doubly intersperse.
Double interspersions yield pairs of ordinary interspersions as defined in [1], and we consider
their fractal sequences.
2. Sequences: Interspersed, Doubly Interspersed
Lemma 1. Suppose that s and x are Fibonacci sequences satisfying
si < xj < si+1 < xj+1 < si+2
(2.1)
for some i and j. Then
si+h < xj+h < si+h+1
for all h ≥ 0 (so that s and x eventually intersperse).
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(2.2)
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THE FIBONACCI QUARTERLY
Proof. For h = 0 and h = 1, the inequalities (2.2) hold by hypothesis. Assume for arbitrary
h ≥ 2 that
si+h−2 < xj+h−2 < si+h−1 and si+h−1 < xj+h−1 < si+h .
Then
si+h−2 + si+h−1 < xi+h−2 + xj+h−1 < si+h−1 + si+h ,
so that (2.2) holds.
¤
Lemma 2. Suppose that s and x are Fibonacci sequences satisfying xk = sk and xk+h = sk+h
for some k ≥ 1 and h ≥ 1. Then x = s.
Proof. Given that xk = sk , if xk+1 = sk+1 , then clearly x = s. On the other hand, if
xk+1 < sk+1 , then
xk+2 = xk + xk+1 < sk + sk+1 = sk+2 ,
and inductively, xk+h < sk+h for all h ≥ 1, contrary to the hypothesis. If sk+1 < xk+1 then
inductively sk+h = xk+h for all h ≥ 1, a contradiction. Therefore, x = s.
¤
Lemma 3. If s is a Fibonacci sequence, then
sn = s1 Fn−3 + s2 Fn−2
for all n ≥ 3, where F0 = 0.
Proof. By induction.
¤
Theorem 1. Distinct positive Fibonacci sequences s and x eventually doubly intersperse if
and only if there exist a ≥ 1 and b ≥ 1 such that
xb − sa = τ (sa+1 − xb+1 ),
(2.3)
√
where τ denotes the golden ratio, (1 + 5)/2. Otherwise, s and x eventually intersperse.
Proof. Suppose that s and x are distinct positive Fibonacci sequences and i and j satisfy
si−1 ≤ si < xj .
(2.4)
The following cases are exhaustive:
Case 1. si < xj < si+1 < xj+1 < si+2 ;
Case 2. si < xj < xj+1 ≤ si+1 ;
Case 3. si < xj < si+1 < si+2 ≤ xj+1 .
In Case 1, the sequences s and x eventually intersperse by Lemma 1. In Case 3, there are
two subcases:
xj−1 ≤ si < xj < si+1 < si+2 ≤ xj+1 ;
(2.5)
si < xj−1 < xj < si+1 < si+2 ≤ xj+1 .
(2.6)
However, for (2.5), the inequalities xj−1 ≤ si and xj < si+1 imply xj+1 < si+2 , a contradiction. For (2.6), clearly si+1 < xj+1 < si+3 . Now if si+3 < xj+2 , then s and x eventually
intersperse by Lemma 1. On the other hand, if
si+1 < xj+1 < xj+2 ≤ si+3 ,
then this possibility is covered by Case 1 (with a shift of indices). Therefore all that remains
is Case 2, which breaks into two cases.
Case 2.1. xj+2 < si+2 ;
Case 2.2. xj+2 ≥ si+2 .
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In Case 2.1, we have si−1 ≤ si < xj and si < xj+1 , so that
si+1 = si−1 + si < xj + xj+1 = xj+2 .
Consequently,
si < xj < xj+1 ≤ si+1 < xj+2 < si+2 ,
which implies
si+1 < xj+2 < si+2 < xj+3 < si+3 ,
so that s and x eventually intersperse by Lemma 1.
In Case 2.2, if there exists h ≥ 1 for which
si+h < xj+h < si+h+1 < xj+h+1 < si+h+2 ,
then s and x eventually intersperse by Lemma 1. In the remaining case that no such h exists,
we have
si < xj < xj+1 ≤ si+1 < si+2 ≤ xj+2 < xj+3 ≤ si+3 < si+4 ≤ · · · .
(2.7)
Although (2.7) shows xj+h ≤ si+h for infinitely many h, at most one of these is xj+h = si+h
by Lemma 2. Likewise, at most one of the inequalities si+k ≤ xj+k in (2.7) is actually
si+k = xj+k Therefore, in (2.7) we can (and do) assume i large enough that
si < xj < xj+1 < si+1 < si+2 < xj+2 < xj+3 < si+3 < si+4 < · · · ,
(2.8)
which is to say that s and x eventually doubly intersperse. From (2.8) we extract two families
of inequalities:
si < xj
si + si+1 = si+2 < xj+2 = xj + xj+1
2si + 3si+1 = si+4 < xj+4 = 2xj + 3xj+1
..
.
Fn si + Fn+1 si+1 < Fn xj + Fn+1 xj+1 , for n = 1, 3, 5, · · ·
and
xj+1 < si+1
xj + 2xj+1 = xj+3 < si+3 = si + 2si+1
3xj + 5xj+1 = xj+5 < si+5 = 3si + 5si+1
..
.
Fn xj + Fn+1 xj+1 < Fn si + Fn+1 si+1 , for n = 2, 4, 6, · · · .
These two families are equivalent to the following pair:
xj − si
Fn+1
<
for n = 1, 3, 5, · · · ;
Fn
si+1 − xj+1
xj − si
Fn+1
>
for n = 2, 4, 6, · · · .
Fn
si+1 − xj+1
Taking limits as n → ∞ results in (2.3). To summarize, if s and x eventually doubly
intersperse, then (2.3) holds. The converse follows from the fact that the steps leading to
(2.3) are reversible.
¤
FEBRUARY 2010
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THE FIBONACCI QUARTERLY
According to Theorem 1, any s and x that do not eventually doubly intersperse must
eventually intersperse, and one should surely then ask “where the interspersion commences.”
A practical way to answer this question for given specific sequences is to determine the least
i for which there is a j satisfying (2.1).
To determine a and b as in (2.3), one can list the first few terms of the sequences (xn − sn ),
(xn+1 − sn ), (xn+2 − sn ), . . ., writing all the terms in the form h − kτ . Then look for a term
h − kτ whose successor is h + k − hτ . The reason this works is that if xb − sa = h − kτ , then
xb − sa = (−τ )(h + k − hτ ) so that h + k − hτ is the desired number xb+1 − sa+1 .
Example 1. Define s1 = 1, s2 = 4, x1 = τ + 1, and x2 = 3, so that
s = (1, 4, 5, 9, 14, 23, . . .),
x = (τ + 1, 3, τ + 4, τ + 7, 2τ + 11, . . .),
and x doubly intersperses s from s2 = 4 and x3 = τ + 4. The necessary and sufficient
condition (2.3) holds for a = b = 1.
Corollary 1. Suppose that s and x eventually doubly intersperse and that a, b are indices
for which (2.3) holds. Then
sa+n − xb+n = (Fn+1 − τ Fn )(sa − xb )
for n = 0, 1, 2, . . ..
Proof. The hypothesis (2.3) is equivalent to
sa+1 − xb+1 = (1 − τ )(sa − xb ).
Then
sa+2 − xb+2 = (sa+1 − xb+1 ) + (sa − xb )
= (F3 − τ F2 )(sa − xb ),
and the proof follows by induction.
¤
Corollary 2. Suppose that s and x eventually doubly intersperse and that a, b are indices
for which (2.3) holds. Then
xb+n = (Fn+1 − τ Fn )xb + Fn (sa+1 + (τ − 1)sa )
(2.9)
for n = 0, 1, 2, . . ..
Proof. By (2.3),
xb+1 = sa+1 − (1/τ )(xb − sa )
= (1 − τ )xb + sa+1 + (τ − 1) sa
= (F2 − τ F1 )xb + F1 (sa+1 + (τ − 1) sa ),
and (2.9) follows by induction.
¤
Example 2. It is easy to construct sequences which eventually doubly intersperse the
Fibonacci numbers. Here we choose to insert x1 and x2 between 5 and 8. Choosing x1 = 6,
we determine the other terms of x by (2.9) using a = b = 2:
s = (5, 8, 13, 21, 34, . . .),
x = (6, 9 − τ, 15 − τ, 24 − 2τ, 39 − 3τ, . . . , Fn+4 + Fn − τ Fn−1, . . .).
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DOUBLY INTERSPERSED AND FRACTAL SEQUENCES
Having succeeded with x1 = 6, it is easy to check that if x1 = 7, there is no choice of x2 for
which (2.8) holds using i = j = 1. In order to find an upper bound for x1 in cases such as
this, we turn to another corollary.
Corollary 3. Suppose that s = (s1 , s2 , . . .) and x = (x1 , x2 . . .) are positive Fibonacci sequences satisfying
s 1 < x 1 < x 2 < s2 < s 3 < x 3 < x 4 < s 4 < s 5 < · · · .
(2.10)
Then
x1 < (2 − τ )s1 + (τ − 1)s2 .
Proof. Taking n = 1, a = 1, and b = 1 in (2.9) gives
(1 − τ )x1 + s2 + (τ − 1)s1 = x2 > x1 ,
so that
x1 <
s2 + (τ − 1)s1
= (2 − τ )s1 + (τ − 1)s2 .
τ
¤
Example 3. Extending Example 2, for s = (5, 8, 13, . . .), we can make a sequence x satisfying
(2.10) (so that x doubly intersperses s from s2 and x3 ) by choosing any x1 in the interval
(5, 3τ + 2) and then applying (2.9). One such sequence is
x = (4τ, 5τ − 1, 9τ − 1, 14τ − 2, 23τ − 3, . . . , (Fn+3 + Fn+2 )τ − Fn−1, . . .).
3. Double Interspersions
A Stolarsky interspersion [3] is an array of positive integers, each occurring exactly once,
such that the rows of the array are Fibonacci sequences, each pair of which eventually
intersperse, and such that the first column of the array is increasing. This last condition
ensures that the first row is the sequence (F2 , F3 , . . .) = (1, 2, 3, 5, 8, . . .) of distinct Fibonacci
numbers. Among Stolarsky interspersions are the original Stolarsky array [8], the Wythoff
array [5, 7], the golden array [2], and the Wythoff dual array [2].
An array of real numbers will be called a double interspersion if every pair of its row sequences eventually doubly intersperse. We shall present a double interspersion R = {R(n, k)}
in which every row is a positive Fibonacci sequence. By (2.3), R cannot consist solely of
integers; however, we shall construct from R an array which is a double interspersion consisting solely of positive integers, although the rows are not Fibonacci sequences. The array
R, defined by
½
Fk+1
if n = 1
R(n, k) =
(3.1)
Fn+k−1 + Fk+2 − τ Fk+1 if n ≥ 2,
begins thus:
1
2
3 − τ 5 − 2τ
4 − τ 6 − 2τ
5 − τ 8 − 2τ
7 − τ 11 − 2τ
10 − τ 16 − 2τ
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3
8 − 3τ
10 − 3τ
13 − 3τ
18 − 3τ
26 − 3τ
5
13 − 5τ
16 − 5τ
21 − 5τ
29 − 5τ
42 − 5τ
8
21 − 8τ
26 − 8τ
34 − 8τ
47 − 8τ
68 − 8τ .
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THE FIBONACCI QUARTERLY
If the numbers in R are jointly ranked in increasing order and then each is replaced by
its rank, the resulting array (A161179 in the Online Encyclopedia of Integer Sequences [6])
begins thus:
1 4 7 12 17 24 31 40 49
2 3 8 11 18 23 32 39 50
5 6 13 16 25 30 41 48 61
9 10 19 22 33 38 51 58 73
14 15 26 29 42 47 62 69 86
20 21 34 37 52 57 74 81 100 .
b we find inductively that
Denoting this order array by R,

(k 2 + 2k − 1)/2
if k is odd and n = 1


 2
(k
+
2k)/2
if
k is even and n = 1
b k) =
R(n,
(3.2)
2
2
[n
+
(2k
−
1)n
+
k
−
2k
−
1]/2
if
k is odd and n ≥ 2


 [n2 + (2k − 3)n + k 2 − 2k]/2
if k is even and n ≥ 2.
b k) are clearly double interspersions which share the same
The arrays R(n, k) and R(n,
commencement function D, defined just after (1.2). This function, one can easily show, is
given by
½
(2, 3)
if (n, k) = (1, 1)
D(n, k) =
(k − n, 1)
otherwise.
4. From Double to Two Singles
b1 and R
b2
An arbitrary double interspersion R necessarily splits into two interspersions R
b1 is the order array of the union of the odd-numbered columns of R,
in the following way: R
b2 is the order array of the union of the even-numbered columns of R after the first two
and R
rows have been swapped. Taking R to be the array (3.1), or equivalently, taking R to the
b in (3.2), we find inductively that
array R

if n = 1
 k2
b1 (n, k) =
k 2 + (n − 2)k + (n2 − 2n + 4)/4 if n is even
R
 2
k + (n − 2)k + (n2 − 2n + 1)/4 if n is odd and n ≥ 3;
b1 :
this array is indexed [6] as A163253. Shown here is a corner of R
1 4 9 16
2 5 10 17
3 7 13 21
6 11 18 27
8 14 22 32
12 19 28 39
25
26
31
38
44
52
36
37
43
51
58
67
49 64 81
50 65 82
57 73 91
66 83 102
74 92 112
84 103 124 .
b1 , beginning with row 5, satisfy a 3rd-order recurrence:
The columns of R
an = an−1 + an−2 − an−3 + 1.
(4.1)
It is noteworthy that the set of non-squares (A000037 in [6]) can be partitioned into a set
b1 , obtaining
of sequences satisfying (4.1). To do so, merely remove the first row from R
A163254, of which the order array is an interspersion, A163255.
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VOLUME 48, NUMBER 1
DOUBLY INTERSPERSED AND FRACTAL SEQUENCES
b2 is given by
The array R

2

 k2 + k − 1

k +k
b2 (n, k) =
R
 k 2 + (n − 1)k + (n − 1)(n − 3)/4

 k 2 + (n − 1)k + n(n − 4)/4
if
if
if
if
n=1
n=2
n is odd and n ≥ 3
n is even and n ≥ 4.
b2 satisfy (4.1). For each n, row n of R
b2 has exactly
Beginning with row 6, the columns of R
b1 .
one term that is also in row n of R
It is proved elsewhere [4] that every interspersion I(n, k) begets a fractal sequence a by
the rule
a(m) = the number n such that m = I(n, k).
b1 is especially simple if broken into segments beginning
The fractal sequence A163256 of R
with 1:
1 2 3
1 2 4 3 5
1 2 4 6 3 5 7
1 2 4 6 8 3 5 7 9
1 2 4 6 8 10 3 5 7 9 11 .
If n > 1, then row n comes from row n − 1 by inserting 2n just before 3 and 2n + 1 at the
end. The resulting sequence
(1, 2, 3, 1, 2, 4, 3, 5, 1, 2, 4, 6, 3, 5, 7, . . .)
is “fractal” in the sense that it contains infinitely many proper subsequences that are identical
b2 and
to the sequence itself. (For the full definition of fractal sequence, see [4]. The array R
its fractal sequence appear in [6] as A163257 and A163258, respectively.)
It is easy to construct other double interspersions and associated interspersions and fractal
sequences. Another particularly simple one, in some sense a dual to R, is given by
½
Fk+1
if n = 1;
T (n, k) =
τ (Fn+k−2 + 1)Fk+2 − Fk+1 if n ≥ 2,
it begins thus:
2τ
3τ
4τ
6τ
9τ
1
2
3
− 2 3τ − 3 5τ − 5
− 1 4τ − 2 7τ − 3
− 1 6τ − 2 10τ − 3
− 1 9τ − 2 15τ − 3
− 1 14τ − 2 23τ − 3
5
8
8τ − 8 13τ − 13
11τ − 5 18τ − 8
16τ − 5 26τ − 8
24τ − 5 39τ − 8
37τ − 5 60τ − 8 .
References
[1] C. Kimberling, Interspersions and dispersions, Proceedings of the American Mathematical Society, 117
(1993), 313–321.
[2] C. Kimberling, Ordering the set of all positive Fibonacci sequences, Applications of Fibonacci Numbers,
Proceedings of the Fifth International Conference on Fibonacci Numbers and Their Applications, Vol. 5,
G. E. Bergum, A. N. Philippou, A. F. Horadam, editors, Kluwer Academic Publishers, Dordrecht, 1993,
405–416.
FEBRUARY 2010
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THE FIBONACCI QUARTERLY
[3] C. Kimberling, Stolarsky interspersions, Ars Combinatoria, 35 (1995), 129–138.
[4] C. Kimberling, Fractal sequences and interspersions, Ars Combinatoria, 45 (1997), 157–168.
[5] D. R. Morrison, A Stolarsky array of Wythoff pairs, A Collection of Manuscripts Related to the Fibonacci
Sequence, Fibonacci Association, Santa Clara, 1980, 134–136.
[6] N. J. A. Sloane, On-Line Encyclopedia of Integer Sequences,
http://www.research.att.com/~njas/sequences/.
[7] N. J. A. Sloane, Classic Sequences, http://www.research.att.com/~njas/sequences/classic.html.
[8] K. B. Stolarsky, A set of generalized Fibonacci sequences such that each natural number belongs to exactly
one, The Fibonacci Quarterly, 15 (1977), 224.
MSC2000: 11B39
Department of Mathematics, University of Evansville, 1800 Lincoln Avenue, Evansville,
IN 47722
E-mail address: [email protected]
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